Question: The equation of hyperbola $H$ is $\dfrac {(x-2)^{2}}{36}-\dfrac{y^2}{49} = 1$. What are the asymptotes?
We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac{y^2}{49} = - 1 + \dfrac {(x-2)^{2}}{36}$ Multiply both sides of the equation by $49$ $y^2 = { - 49 + \dfrac{ (x-2)^{2} \cdot 49 }{36}}$ Take the square root of both sides. $\sqrt{y^2} = \pm \sqrt { - 49 + \dfrac{ (x-2)^{2} \cdot 49 }{36}}$ $ y = \pm \sqrt { - 49 + \dfrac{ (x-2)^{2} \cdot 49 }{36}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y \approx \pm \sqrt {\dfrac{ (x-2)^{2} \cdot 49 }{36}}$ $y \approx \pm \left(\dfrac{7 \cdot (x - 2)}{6}\right)$ Rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{7}{6}(x - 2)+ 0$